Problem: $\begin{aligned} &f(a)=3(a+2)^2-10 \\\\ &g(b)=\dfrac{2}{5}b+4 \end{aligned}$ $g(f(-2))=$
Answer: When evaluating composite functions, we work our way inside out. To evaluate $g(f(-2))$, let's first evaluate $f(-2)$. Then we'll plug that result into $g$ to find our answer. Let's evaluate $f({-2})$. $\begin{aligned}f(a)&=3(a+2)^2-10\\\\ f({-2})&=3(({-2})+2)^2-10 ~~~~~~~~~~\text{Plug in }a={-2}\\\\ &=3(0)^2-10\\\\ &={-10}\end{aligned}$ We now know that $g(f({-2}))$ is the same as $g({-10})$ because $f({-2}) = {-10}$. Let's evaluate $g({-10})$. $\begin{aligned}g(b)&=\dfrac{2}{5}b+4\\\\ g({{-10}})&=\dfrac{2}{5}({-10})+4~~~~~~~~~~\text{Plug in }b={-10}\\\\ &=-4+4\\\\ &=0\\\\\end{aligned}$ The answer: $g(f(-2))= 0$